Let A=[1+x2-y2-z2 2(x y+z) 2(z x-y) 2(x y-z) 1+y2-z2-x2 2(y z+x) 2(z x+y) 2(y z-x) 1+z2-x2-y2] then det. A is equal to

Question

Let A=[1+x2-y2-z2 2(x y+z) 2(z x-y) 2(x y-z) 1+y2-z2-x2 2(y z+x) 2(z x+y) 2(y z-x) 1+z2-x2-y2] then det. A is equal to (A) (1+x y+y z+z x)3 (B) (1+x2+y2+z2)3 (C) (x y+y z+z x)3 (D) (1+x3+y3+z3)2

Tardigrade Admin · Accepted Answer

multiply R2 by z and R3 by y and use R1 arrow R1-R2+R3 Objective approach: put z = y =0 then choices are A =1 ; B =(1+ x 2)3 ; C =0 ; D =(1+ x 3)2 and determinant comes out to be (1+x2)3 ⇒(B)

Q. Let $A = ⎣ ⎡ 1 + x^{2} - y^{2} - z^{2} 2 (x y - z) 2 (z x + y) 2 (x y + z) 1 + y^{2} - z^{2} - x^{2} 2 (yz - x) 2 (z x - y) 2 (yz + x) 1 + z^{2} - x^{2} - y^{2} ⎦ ⎤$ then det. $A$ is equal to

$(1 + x y + yz + z x)^{3}$

$(1 + x^{2} + y^{2} + z^{2})^{3}$

$(x y + yz + z x)^{3}$

$(1 + x^{3} + y^{3} + z^{3})^{2}$

multiply $R_{2}$ by $z$ and $R_{3}$ by y and use $R_{1} \to R_{1} - R_{2} + R_{3}$
Objective approach : put $z = y = 0$ then choices are $A = 1; B = (1 + x^{2})^{3}; C = 0; D = (1 + x^{3})^{2}$ and determinant comes out to be $(1 + x^{2})^{3} \Rightarrow (B)$

Q. Let A=⎣⎡​1+x2−y2−z22(xy−z)2(zx+y)​2(xy+z)1+y2−z2−x22(yz−x)​2(zx−y)2(yz+x)1+z2−x2−y2​⎦⎤​ then det. A is equal to

(1+xy+yz+zx)3

(1+x2+y2+z2)3

(xy+yz+zx)3

(1+x3+y3+z3)2